# Homework Help: Multiplying a fraction

1. Dec 24, 2015

### science_rules

1. The problem statement, all variables and given/known data
I would just like to know if I got this right. ((x2 + 2x + 1)/(18cw3) times ((12c3w)/(x2 - 1)) = (12c3wx2 + 2x12c3w + 12c3w)/(18cw3)(x2 - 18cw3

2. Relevant equations
((x2 + 2x + 1)/(18cw3) times ((12c3w)(x2 - 1))

3. The attempt at a solution
(12c3wx2 + 2x12c3w + 12c3w)/((18cw3)(x2 - 1)) = (12c3wx2 + 2x12c3w + 12c3w)/(18cw3x2 - 18cw3

2. Dec 24, 2015

### Samy_A

Removed, as apparently there is an error in the formulation of the question.

3. Dec 24, 2015

### science_rules

I accidentally used a subscript all the way through instead of a power, argh!!

4. Dec 24, 2015

### Staff: Mentor

On the right side one of the terms is $2x12c_3w$. Should that have been $2x^{12}c^3$? Also, you're missing a right parenthesis.

5. Dec 25, 2015

### haruspex

Do you mean $\frac{x^2+2x+1}{18cw^3}\frac{12c^3w}{x^2-1}$?
If so, there is a lot of cancellation available. Factorise the two expressions involving x.

6. Dec 25, 2015

### science_rules

It should have been as haruspex said: (x2 + 2x + 1)/(18cw3) times (12c3w)/(x2-1)

7. Dec 25, 2015

### science_rules

Yes, that is what I meant.

8. Dec 25, 2015

### science_rules

(x+1)(x+1)/(18cw3) times (12c3w)/(x-1)(x+1)
The x+1 cancels, leaving: (x+1)/(18cw3) times (12c3w)/(x-1)

9. Dec 25, 2015

### haruspex

More cancellation to go. Look at the constants, look at the powers of c and w.

10. Dec 25, 2015

### science_rules

(x+1)/((3w)(6cw2)) times ((3w)(4c3))/(x-1) = (x+1)/(6cw2) times (4c3)/(x-1)

11. Dec 25, 2015

### science_rules

The 3w is canceled

12. Dec 25, 2015

### science_rules

I meant that 3wc is canceled, leaving: (x+1)/(w2) times (4c2)/(x-1)

13. Dec 25, 2015

### haruspex

You made a mistake with the constants.

14. Dec 25, 2015

### science_rules

Should it be: 3cw(w2) and 3cw(4c2)??

15. Dec 25, 2015

### science_rules

Oops I meant: 3cw(6w2) and 3cw(4c2)

16. Dec 25, 2015

### haruspex

Yes.

17. Dec 25, 2015

### science_rules

Thank you for your help

18. Dec 25, 2015

### science_rules

So the answer is: (x+1)/(6w2) times (4c2)/(x-1)

19. Dec 26, 2015

### Staff: Mentor

Instead of adding new posts each time you discover something you should have said, you can edit the post you want to change.

20. Dec 26, 2015

### Thewindyfan

The constants can be further simplified, and then you should bring it together so you have one fraction.

21. Dec 26, 2015

### science_rules

Like this?: (x+1)/(6w2) times (4c2)/(x-1) = 4c2x + 4c2/6w2x-6w2 = 2c(2cx+2c)/2w(3wx-3w)<------(Answer)?

22. Dec 27, 2015

### Samy_A

1)
2) I think you can leave the nominator as $2c²(x+1)$, and similarly for the denominator. Certainly would give a better readable result.

Last edited: Dec 27, 2015
23. Dec 27, 2015

### science_rules

How did you get 2c2(x+1) when it was 4c2/(x+1)?? Isn't the numerator supposed to be 2c2(2x+2)??
And the denominator is supposed to be: 2w2(3x-3)??

24. Dec 27, 2015

### haruspex

You are still not doing all the simplification you could. In what you posted above, each parenthetic term has a factor that can be taken outside. Having done that, one of the primes occurs above and below the line.

25. Dec 27, 2015

### science_rules

Okay, I think I have it now: (x2 + 2x + 1)/(18cw3) times (12c3w)/(x2-1) = (x+1)2(12c3w)/(18cw2)(x+1)(x-1) Cancel the (x+1) = (12c3w)(x+1)/(18cw2)(x-1) Cancel the c and w = (12c2)(x+1)/(18w)(x-1) divided by 3 (numerator and denominator) = (4c2)(x+1)/(6w)(x-1) divided by 2 (numerator and denominator) = (2c2)(x+1)/(3w)(x-1) Answer